\operatorname { sin } x + \operatorname { sin } 2 x + \operatorname { sin } 4 x + \operatorname { sin } 5 x = 4 \operatorname { cos } \frac { x } { 2 } \operatorname { cos } \frac { 3 x } { 2 } \operatorname { sin } 3 x

\operatorname { sin } x + \operatorname { sin } 2 x + \operatorname {

1.	\operatorname { sin } x + \operatorname { sin } 2 x + \operatorname { sin } 4 x + \operatorname { sin } 5 x = 4 \operatorname { cos } \frac { x } { 2 } \operatorname { cos } \frac { 3 x } { 2 } \operatorname { sin } 3 x
Answer» Sinx+sin2x+sin4x+sin5x=2sin(x+2x)/2cos(x-2x)/2+2sin(4x+5x)/2xos(4x-5x)/2[sinC+sinD=2sin(C+D)/2cos(C-D)/2]=2sin3x/2cosx/2+2sin9x/2cosx/2 [∵, cos(-x)=cosx]=2cosx/2(sin3x/2+sin9x/2)=2cosx/2{2sin(3x/2+9x/2)/2cos(3x/2-9x/2)/2}=2cosx/2(2sin12x/4cos6x/4)=4cosx/2sin3xcos3x/2=4cosx/2cos3x/2sin3x (Proved)